A loop of wire (resistance = 2.0 mΩ)is positioned as shown with respect to a long wire which carries a current.If d = 1.0 cm,D = 6.0 cm,and L = 1.5 m,what current is induced in the loop at an instant when the current in the wire is increasing at a rate of 100 A/s?

Question

Quizplus · Accepted Answer

The induced current in the loop is given by the equation:

I = (μ0/4π) * (2πI1R^2 / (R^2 + (L/2)^2)^3/2) * (dI2/dt)

where μ0 is the permeability of free space, I1 is the current in the long wire, R is the distance from the wire to the center of the loop, L is the length of the wire, and dI2/dt is the rate of change of current in the wire.

Substituting the given values, we get:

I = (4π*10^-7 Tm/A) * (2π* I1 * (0.06)^2 / ((0.06)^2 + (1.5/2)^2)^3/2) * 100 A/s

I = 26.9 mA

Therefore, the best choice is option C.

A Loop of Wire (Resistance = 2